23 11
Article 09.1.6
Journal of Integer Sequences, Vol. 12 (2009), 2
3 6 1 47
On the Expansion of Fibonacci and Lucas
Polynomials
Helmut Prodinger
Department of Mathematics
University of Stellenbosch
7602 Stellenbosch
South Africa
hproding@sun.ac.za
Abstract
Recently, Belbachir and Bencherif have expanded Fibonacci and Lucas polynomials using bases of Fibonacci- and Lucas-like polynomials. Here, we provide simplified proofs for the expansion formulæthat in essence a computer can do. Furthermore, for 2 of the 5 instances, we findq-analogues.
1
Introduction
In [2], Belbachir and Bencherif studied the Fibonacci and Lucas polynomials:
U0 = 0, U1 = 1, Un =xUn−1+yUn−2,
V0 = 2, V1 =x, Vn =xVn−1+yVn−2.
We prefer the modified polynomials
u0 = 0, u1 = 1, un=un−1+zun−2,
v0 = 2, v1 = 1, vn=vn−1+zvn−2,
so that
Un(x, y) =xn−1un
y
x2
, Vn(x, y) = xnvn
y
x2
.
Then, with
λ1,2 =
1±√1 + 4z
un =
Substitutingz =t/(1−t)2, these formulæ become particularly nice:
un=
1−(−t)n
(1 +t)(1−t)n−1, vn=
1 + (−t)n
(1−t)n .
The main result of [2] are the following 5 formulæ:
2u2n+1=
But the proofs of all these, using the simple forms forun andvn, can be done by a computer!
To give the reader an idea, let us do the last one, which seems to be the most complicated:
n
The other proofs are similar/easier:
n
X
k=1
cn,ku2n−k =
2(1−t2n)
(1−t)2n−1(1 +t)−
1 +t2n−1
(1 +t)(1−t)2n−2
= 1−t
2n−1
(1−t)2n−1 =v2n−1.
Remark 1. The polynomials un(x) and vn(x) are essentially Chebyshev polynomials.
The authors of [2] have also published a companion paper [3]; according to a remark in [2], the results in [2] are more general than the ones in [3].
2
q
-analogues
Now we are interested in q-analogues. For this, we replace un by
Fibn =
X
0≤k≤n−1
2
q(k+12 )
n−k−1
k
q
zk
and vn by
Lucn =
X
0≤k≤n2 q(k2)
n−k k
q
[n]q
[n−k]q
zk,
as suggested by Cigler [4]. We use standard q-notation here:
[n]q := 1 +q+· · ·+qn−1, [n]q! := [1]q[2]q. . .[n]q,
n k
q
:= [n]q! [k]q![n−k]q!
,
compare [1]; the notions of the Introduction are the special instanceq = 1.
Theorem 2.
Luc2n−1 =
n
X
k=1
dn,kLuc2n−1−k,
with
dn,k = (−1)k−1
q(k2)
1 +qn−1
n−1
k
q
+qn−1
n k
q
.
Proof. We must prove that
X
0≤k≤n−1
q(k2)
2n−1−k k
q
[2n−1]q
[2n−1−k]q
zk
=
n
X
j=1
(−1)j−1 q(
j 2)
1 +qn−1
n−1
j
q
+qn−1
n j
q
× X
0≤k≤2n−j−1
2
q(k2)
2n−j −1−k k
q
[2n−j−1]q
[2n−j−1−k]q
Comparing coefficients, we have to prove that
Simplifying, we must prove that
n
Another form of this is
n
We write the desired identity as
n
can be shown by inspection, and for k =n, the identity holds, since the sum is empty.
Proof. We must prove that
X
0≤k≤n−1
q(k+12 )
2n−k−1
k
q
zk
=
n
X
j=1
(−1)j−1q(j2)
n j
q
X
0≤k≤2n−j−1
2
q(k+12 )
2n−j −k−1
k
q
zk.
Comparing coefficients, this means
n
X
j=0
(−1)jq(j2)
n j
q
2n−j−k−1
k
q
= 0,
which follows by a similar but simpler argument than before.
3
Conclusion
We found 2q-analogues; for the remaining 3 instances we were not successful and leave this as a challenge for anybody who is interested.
References
[1] G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 2000.
[2] H. Belbachir and F. Bencherif, On some properties of bivariate Fibonacci and Lucas
polynomials,J. Integer Sequences 11 (2008), Article 08.2.6.
[3] H. Belbachir and F. Bencherif, On some properties of Chebyshev polynomials,Discuss. Math. Gen. Algebra Appl. 28 (2008), 121–133.
[4] J. Cigler, A new class of q-Fibonacci polynomials, Electronic J. Combinatorics 10
(2003), Article R19. Text available at
http://www.combinatorics.org/Volume 10/Abstracts/v10i1r19.html.
2000 Mathematics Subject Classification: Primary 11B39; Secondary 11B37.
Keywords: Fibonacci polynomials, Lucas polynomials, generating functions, q-analogues.
(Concerned with sequences A007318, A029653and A112468.)
Received October 6 2008; revised version received December 15 2008. Published inJournal of Integer Sequences, December 17 2008.
